2002 AMC 10A Problem 11

This is a "smallest number" question, so it has two halves: show that $13$ disks are enough, then show that $12$ are not.

Setting up. There are $3$ files of $0.8$, $12$ files of $0.7$, and $30 - 15 = 15$ files of $0.4$. Call the $0.8$ and $0.7$ files big ($15$ of them) and the $0.4$ files small ($15$ of them).

1. What fits on one disk ($1.44$ MB)?
- Two big files: only $0.7 + 0.7 = 1.4$ fits ($0.8 + 0.7 = 1.5$ is too much), and then there is no room for anything else.
- One big file: it fits with one small file ($0.8 + 0.4 = 1.2$), but not two ($0.7 + 0.4 + 0.4 = 1.5$).
- No big files: at most three small files ($1.2$), since four would be $1.6$.

Part 1: thirteen disks are enough.

1. Put each $0.8$ file with one $0.4$ file: $3$ disks. Pair the twelve $0.7$ files: $6$ disks. The $15 - 3 = 12$ small files left go three to a disk: $4$ disks. That is $3 + 6 + 4 = 13$ disks.

Part 2: twelve disks are not enough.

3. Give every big file $3$ points and every small file $2$ points. Using the list in step 1, the most points a disk can carry is:
- two big files: $3 + 3 = 6$,
- one big and one small: $3 + 2 = 5$,
- three small: $2 + 2 + 2 = 6$.

So every disk holds at most $6$ points.

1. The files are worth $15 \times 3 + 15 \times 2 = 45 + 30 = 75$ points in total. Twelve disks carry at most $12 \times 6 = 72$ points, which is less than $75$. So $12$ disks (or fewer) cannot hold everything.

Thirteen disks work and twelve do not, so the smallest number is $13$.

The answer is $\boxed{\textbf{(B)}\ 13}$.

Adding up megabytes gives a lower bound that is too weak, because files cannot be split and space is wasted. Giving each kind of file a score, chosen so that a full disk is worth at most a fixed amount, measures the waste correctly. For any "smallest number needed" question, always do both halves: an arrangement that works, and a reason nothing smaller can.

Handle the $0.8$ files first. Each one needs its own disk, and the only thing that fits beside it is one $0.4$ file, so use $3$ disks for $0.8 + 0.4$. Left over are twelve $0.7$ files and twelve $0.4$ files, totalling $12 \times 1.1 = 13.2$ MB. Nine disks hold only $9 \times 1.44 = 12.96$ MB, so at least $10$ more disks are needed, and six disks of $0.7 + 0.7$ plus four disks of three $0.4$ files do it. Total: $13$.

There are $3$ files of $0.8$, $12$ of $0.7$, and $30 - 15 = 15$ of $0.4$. Call the $0.8$ and $0.7$ files "large" ($15$ in all). Check what fits in $1.44$: two large files fit only if both are $0.7$ ($1.4$), and then nothing else fits; one large file leaves room for at most one $0.4$ ($0.7 + 0.4 + 0.4 = 1.5$ is too much); a disk with no large file holds at most three $0.4$ files ($1.2$).

Suppose $12$ disks were enough. Let $x, y, z$ be the numbers of disks holding two, one, or zero large files. Then $x + y + z = 12$ and $2x + y = 15$, so $x = 3 + z$ and $y = 9 - 2z$. Those disks can hold at most $y + 3z = 9 + z$ small files, and $y \ge 0$ forces $z \le 4$, so at most $13 < 15$ small files fit. Contradiction: $12$ disks fail.

Thirteen disks work: three disks each with $0.8 + 0.4$, six disks each with $0.7 + 0.7$, and four disks each with three $0.4$ files. That stores $3 + 12 + 15 = 30$ files.

The answer is $\boxed{\textbf{(B)}\ 13}$.